A uniform rod of mass m, length l is placed on smooth horizontal surface. An impulse J is given horizontally at one end perpendicular to the length of rod. Kinetic energy of the rod after impulse is given, is
Question
A uniform rod of mass m, length l is placed on smooth horizontal surface. An impulse J is given horizontally at one end perpendicular to the length of rod. Kinetic energy of the rod after impulse is given, is
Solution
The kinetic energy of the rod after the impulse can be calculated using the principles of conservation of linear and angular momentum.
Step 1: Calculate the linear velocity of the center of mass. The impulse given to the rod will cause it to move. The velocity (v) of the center of mass of the rod can be calculated using the formula for impulse, which is the product of the mass and the change in velocity. So, v = J/m.
Step 2: Calculate the angular velocity of the rod. The impulse will also cause the rod to rotate. The angular velocity (ω) can be calculated using the formula for the moment of impulse, which is the product of the impulse and the distance from the point of application to the center of mass (l/2 in this case), divided by the moment of inertia of the rod. The moment of inertia of a rod about its center of mass is (ml^2)/12. So, ω = J(l/2)/((m*l^2)/12) = 6J/ml.
Step 3: Calculate the kinetic energy of the rod. The kinetic energy (K) of the rod is the sum of the translational kinetic energy and the rotational kinetic energy. The translational kinetic energy is (1/2)mv^2 and the rotational kinetic energy is (1/2)Iω^2. Substituting the values of v and ω from steps 1 and 2, we get K = (1/2)m(J/m)^2 + (1/2)((ml^2)/12)*(6J/ml)^2 = J^2/(2m) + J^2/(4m) = 3J^2/(4m).
Similar Questions
A uniform rod of mass M and length ℓ is moving on a smooth horizontal plane, such that its one end is moving with a velocity v0 and other end is moving with a velocity 2v0 in the same direction as shown. Then, the kinetic energy of the rod is
1. The bent rod has a mass p per unit length and rotates about the z-axis with an angularvelocity ur. Determine the angular momentum L of the rod about the fixed origin of theaxes, which are attached to the rod. Also find the kinetic energy T of the rod I KineticEnergy is given by: I=ll2 a. L I (the dot product of o and L)
Uniform rod of mass m and length a,P point is at distance a from near end,and at extension of the rod.
The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is 2.00 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.(a) Determine the speed of its center of gravity at its lowest position. m/s(b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod.
The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is 2.00 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.(a) Determine the speed of its center of gravity at its lowest position. Consider the conservation of energy of the center of mass of the rod. What is the relationship between angular and tangential velocities? m/s
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.